Its known that within the perspective of $\sf ZF$ related theories, Scott's definition of cardinality can work under weaker grounds than Von Neumann's! Scott's cardinality works as long as the principle "every set is equinumerous to some well founded set" is in action. Now choice implies that, so it works under choice (even in absence of regularity!), and it can also work in absence of choice and regularity but provided that the above principle is axiomatized.

Can we have a definition of cardinality that is weaker than that of Scott's? That is, it works under all grounds that Scott's cardinality works under, but it can also work at some grounds where Scott's cardinality fails!

Scott's cardinality of a set $x$ is the set of all sets equinumerous to $x$ of the lowest possible rank.

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    $\begingroup$ You do understand that MO is not a platform for publishing new results in this manner, right? $\endgroup$
    – Asaf Karagila
    Commented Jun 30, 2020 at 9:52
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    $\begingroup$ No. I don't understand that. Because it's written that we can ask questions and answer them at the same time, and the question is mathematical. So where did I break the rules? $\endgroup$ Commented Jun 30, 2020 at 9:53
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    $\begingroup$ While the stackexchange technology allows it, I take it to be a social norm of MathOverflow that it is ok to ask a question before one knows the answer, then if later it turns out the asker gets the answer they should answer their own question. Asking and answering in one go is so uncommon, though, that this is the only example I can recall in a decade of using the site. $\endgroup$
    – David Roberts
    Commented Jul 1, 2020 at 4:27

1 Answer 1


The answer is YES!

Let $\mathcal H_\alpha$ stand for the set of all sets hereditarily strictly subnumerous to ordinal $\alpha$.

Now for any set $x$, $\mathcal H^x_{min}$ is meant to be the minimal $\mathcal H_\alpha$ such that there exists an iterative power of it that is supernumerous to $x$. Formally:

Define: $\mathcal H^x_{min} = min \ \mathcal H_\alpha: \exists \beta [ x \rightarrowtail P_\beta(\mathcal H_\alpha)]$

Where $``\rightarrowtail"$ signify "is injective to".

$P_\beta$ is defined recursively as:

$P_\emptyset(x)=x \\ P_{\beta+1}(x)= P(P_\beta(x)) \\ P_\beta(x) = \bigcup (\{P_\alpha(x):\alpha < \beta\}) \text{ if } \beta \text{ is a limit ordinal} $

Now by $P^x_{min}(S) $ its meant the minimal iterative power of $S$ that is supernumerous to $x$. Formally:

Define: $P^x_{min} (S) = min \ P_\beta (S): x \rightarrowtail P_\beta(S) $

Now we come to define cardinality of a set $x$, denoted by $``|x|"$, as the set of all subsets of $P^x_{min} (\mathcal H^x_{min})$ , that are equinumerous to $x$. Formally:

Define: $|x|= \{y| \ y \sim x \land y \subseteq P^x_{min} (\mathcal H^x_{min}) \}$

Where $\sim$ signify "is bijective to"

This definition of cardinality can work under grounds weaker than those of Scott's cardinality? The latter demands the assumption that "every set is equinumerous to some well founded set", and under that assumption the cardinality [defined here] of any set would be exactly its Scott's cardinal. But this definition can work even when the above assumption fails, but it requires the statement: $$\forall x \exists \alpha \exists \beta : x \rightarrowtail P_\beta (\mathcal H_\alpha)$$ which doesn't imply the above assumption! (See this answer, and this).

Note: by well founded set its meant a set whose transitive closure is well founded with respect to $\in$.


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